在 $t$ 小时后，细菌数量 $P$ 满足模型 $P = P_0(2^t)$，其中 $P_0$ 是一个常数。

1. Set up the equation for the population reaching 100 times its original size

Let the original population be $P_0$.
We want to find $t$ when $P = 100P_0$.
$100P_0 = P_0(2^t)$
$2^t = 100$

2. Solve for $t$ using logarithms and round to the nearest hour

Take logarithms on both sides:
$\log(2^t) = \log(100)$
$t \log 2 = 2$
$t = \frac{2}{\log 2} \approx 6.6439$
Since we need the least number of hours to the nearest hour for the population to reach at least 100 times, we round up to 7 hours.

1. Identify the shape and domain of the graph

The function $P = P_0(2^t)$ represents exponential growth since the base $2 > 1$. Since $t$ represents time, the domain is $t \ge 0$.

2. Determine the $P$-intercept and sketch the curve

When $t = 0$, $P = P_0(2^0) = P_0$. The graph starts at $(0, P_0)$ on the $P$-axis and curves upwards with an increasing gradient.